Real masses and squashed spheres

A mass-deformed partition function is useful only after “mass” has a definite meaning. Which flavor charges feel it? Which auxiliary source accompanies it? Likewise, a squashed metric alone does not specify a supersymmetric background. We will supply these data and assemble an integral whose round, massless limit is already known.

Use the first exact partition-function check, the squashed-sphere backgrounds, and real masses from flavor multiplets. Keep compact equal-rank $U(N)_k\times U(N)_{-k}$, integer $k>0$, with the ordinary fundamental traces and no additional gauge quotient. The undeformed flat theory has twelve real Poincaré supercharges, or sixteen at $k=1,2$. Our construction uses its manifest three-dimensional $\mathcal N=2$ description; the sources below establish a specified rigid subalgebra, not preservation of that full extended symmetry.

Choose a complete ellipsoid background

Let $L>0$ be a length scale and $b>0$ a dimensionless deformation. Define $\ell_1=Lb$, $\ell_2=L/b$, and coordinates $0\leq\chi\leq\pi/2$, $\varphi,\vartheta\sim\varphi,\vartheta+2\pi$. Choose

$$\begin{aligned} ds^2&=f(\chi)^2d\chi^2+ \ell_1^2\cos^2\chi\,d\varphi^2+ \ell_2^2\sin^2\chi\,d\vartheta^2,\\ f^2&=\ell_1^2\sin^2\chi+\ell_2^2\cos^2\chi,\\ H&=\frac{i}{f},\qquad V=C=0,\\ A^{(R)}&=\frac12\left(1-\frac{\ell_1}{f}\right)d\varphi+ \frac12\left(1-\frac{\ell_2}{f}\right)d\vartheta . \end{aligned} \tag{AB14.1}$$

This is the toric ellipsoid representative of Willett (2.33), (2.41). It is not the homogeneous, four-supercharge metric also discussed in the prerequisite. Near $\chi=0$ the $\vartheta$ circle collapses and $f(0)=\ell_2$; near $\pi/2$ the $\varphi$ circle collapses and $f(\pi/2)=\ell_1$. These conditions make both the metric and the displayed R connection smooth. The paper’s endpoint labels accompanying its metric are interchanged; we use the directly checked conditions.

Take the orientation of $e^1=f\,d\chi$, $e^2=\ell_1\cos\chi\,d\varphi$, $e^3=\ell_2\sin\chi\,d\vartheta$. The space is the unquotiented $S^3$, with its unique spin structure and trivial R line bundle. The two-form source is specified by $dB=-iH\,d\mathrm{vol}=d\mathrm{vol}/f$ and ordinary patch potentials; its nonzero integral prevents a single global potential. No additional compact two-form flux condition is imposed.

There is one displayed Killing-spinor direction of each R charge. For an explicit check in this diagonal frame, let $\gamma_a$ be the inherited Euclidean matrices, choose constant columns $\gamma_2v_\pm=\pm v_\pm$, and set

$$\zeta=e^{i(\varphi+\vartheta)/2}e^{-i\chi\gamma_1/2}v_-, \qquad \widetilde\zeta=e^{-i(\varphi+\vartheta)/2}e^{-i\chi\gamma_1/2}v_+ . \tag{AB14.2}$$

The angular signs under a $2\pi$ turn are those of the rotating spin frame, not a new antiperiodic boundary condition on the sphere. The smooth spin lift gives global spinors. For example, $\Gamma_\varphi=i\ell_1\sin\chi\,\gamma_3/(2f)$, and the first angular Killing equation reduces to $(1+\cos\chi\,\gamma_2+\sin\chi\,\gamma_3)\zeta=0$. Equation (AB14.2) satisfies it because conjugating $\gamma_2$ by $e^{-i\chi\gamma_1/2}$ gives the combination in parentheses. The radial equation and the opposite-charge equation work in the same frame.

At $b=1$, $f=L$ and $A^{(R)}=0$, but $H=+i/L$. This is the positive auxiliary branch. The previous ABJM integral used $H=-i/L$. We will account for the resulting phase rather than identify these source configurations.

Give a flavor symmetry a real scalar source

The four chirals $A_1,A_2$ and $B_1,B_2$ all retain scalar R-charge $1/2$. Choose one flavor generator with charges

Chiral	Gauge weight on a Cartan background	Flavor charge
$A_1$	$\sigma_i-\widehat\sigma_j$	$+1$
$A_2$	$\sigma_i-\widehat\sigma_j$	$-1$
$B_1$	$-\sigma_i+\widehat\sigma_j$	$-1$
$B_2$	$-\sigma_i+\widehat\sigma_j$	$+1$

Each word in the fixed quartic superpotential contains all four fields, so its total flavor charge is zero. This is a real-mass deformation of a preserved flavor symmetry, not a new quadratic superpotential. We use the compact $U(1)$ cover acting with these integer charges and a unit-charge background probe. This specifies the source normalization without claiming a classification of the faithful flavor group after gauge-center identifications.

Freeze the flavor vector multiplet at

$$a_F=0,\qquad \sigma_F=\mu\in\mathbb R,\qquad D_F=-H\mu=-\frac{i\mu}{f},\qquad \lambda_F=\widetilde\lambda_F=0. \tag{AB14.3}$$

Its gaugino variations vanish: $\partial\mu=0$, $da_F=V=0$ and $D_F+H\mu=0$. This is P (6.26), sufficient for both retained labels even though this geometry has fewer generic solutions than the round sphere. The imaginary, position-dependent $D_F$ is a fixed source, not an integration variable to minimize.

Set bare central parameters to zero. On the dynamical Cartan locus, a field of flavor charge $q_F$ has signed flat-space fermion mass $w=\sigma_i-\widehat\sigma_j+q_F\mu$ for an $A$ field, with the gauge weight reversed for a $B$ field. The P central-charge combination is $-w$, not $w$. At R-charge $1/2$, its scalar fluctuation operator is

$$\mathcal K_B=-\mathscr D^2+w^2-\frac{iw}{f}-\frac{R(g)}8, \qquad \mathcal K_F=\gamma^\mu\mathscr D_\mu+w \tag{AB14.4}$$

in the independent fermion-row convention of the mass lesson. Here $\mathscr D$ includes the R connection with the component’s R charge; the paper’s curvature sign gives $R=-6/L^2$ at the round point. The linear term $-iw/f$ follows from $-2(r-1)H(-w)$ in P (6.32). It would be missed by adding only $w^2$ to a massless scalar operator. At large $L$ with physical $\mu$ fixed, the curvature terms disappear locally and the flat real mass remains.

The dimensionless quantities entering the integral are

$$s_i=L\sigma_i,\qquad t_i=L\widehat\sigma_i,\qquad M=L\mu, \qquad Q_b=b+b^{-1}. \tag{AB14.5}$$

The symbol $Q_b$ is a number, not a supercharge. Holding $M$ fixed while changing $L$ changes the dimensional mass.

Assemble the deformed integral

First evaluate the classical term. The unchanged supersymmetric CS density is $k(i\,a\wedge da-2D\sigma\,d\mathrm{vol}+\cdots)/(4\pi)$. The locus has $a=0$, constant real $\sigma$, and $D=-i\sigma/f$. Since

$$\int_{S^3}\frac{d\mathrm{vol}}f =(2\pi)^2\ell_1\ell_2\int_0^{\pi/2}\sin\chi\cos\chi\,d\chi =2\pi^2L^2,$$

its first-node action is $i\pi k\sum_i s_i^2$. The weight $e^{-S_E}$ is therefore $e^{-i\pi k\sum_i(s_i^2-t_i^2)}$. The change from the earlier positive first-node phase came from $H$ and the locus; the coefficient of the gauge CS action was not changed.

The two root factors now involve both circle scales. Define

$$\mathcal V_b(s)= \prod_{i<j}4\sinh\!\bigl(\pi b(s_i-s_j)\bigr) \sinh\!\bigl(\pi b^{-1}(s_i-s_j)\bigr). \tag{AB14.6}$$

For matter we need the double sine, a regularized product that packages the unpaired modes:

$$s_b(z)=\prod_{m,n\geq0}^{\mathrm{reg}} \frac{mb+nb^{-1}+Q_b/2-iz}{mb+nb^{-1}+Q_b/2+iz}, \qquad D_b(x)=s_b(iQ_b/4+x)s_b(iQ_b/4-x). \tag{AB14.7}$$

The regularization is part of this definition; an ordinary product without subtractions diverges. We use the normalization of Kubo–Nosaka–Pang, Appendix A, including $s_b(z)s_b(-z)=1$. Thus $D_b(x)=s_b(x+iQ_b/4)/s_b(x-iQ_b/4)$ is even in $x$.

For the positive-$H$ branch, a single R-charge-$1/2$ chiral of mass $w$ contributes $s_b(iQ_b/4-Lw)$, as in Willett (3.54). Opposite gauge and flavor charges pair $A_1$ with $B_1$, giving $D_b(s_i-t_j+M)$; the other pair gives $D_b(s_i-t_j-M)$. This single-chiral mass sign is the continuation of (AB14.4); reversing the round auxiliary branch conjugates the real-mass determinant. The paired expression does not depend on interchanging its two signs.

The result is

$$\boxed{\begin{aligned} Z_b(N,k;M)={}&\frac1{(N!)^2} \int_{\mathbb R^{2N}}\prod_i ds_i\,dt_i\, e^{-i\pi k\sum_i(s_i^2-t_i^2)}\\ &\times\mathcal V_b(s)\mathcal V_b(t) \prod_{i,j}D_b(s_i-t_j+M)D_b(s_i-t_j-M). \end{aligned}} \tag{AB14.8}$$

Both Weyl divisions and both scalar centers remain. There is no additional Vandermonde, bosonic level shift or FI parameter. The real Cartan contour and shifted imaginary auxiliary contour are the continuation used in the localization lesson. At finite positive Gaussian damping $e^{-\epsilon\sum_i(s_i^2+t_i^2)}$, the real integral is absolutely convergent. The limiting prescription is the corresponding Abel limit; our rank-one check below does not prove its existence for every rank.

Equation (AB14.8) is the specialization of KNP (2.1)–(2.2) with their integration variables $2\pi s_i,2\pi t_i$ and $(m_1,m_2,m_3)=(0,2M,0)$, after permuting the four factors. The geometry and flavor sources above explain what that specialization means.

Our source prescription is the displayed double-sine normalization and measure, with no additional integer flavor, mixed or gravitational counterterms. This does not assert that all regulator-fixed contact coefficients vanish. Adding a permitted local source action can change the phase; exercise 3 computes one such change.

Recover the round answer and make a mass check

The double-sine shift identity gives

$$D_1(x)=\frac1{2\cosh\pi x}. \tag{AB14.9}$$

At $b=1,M=0$, the root and matter factors in (AB14.8) reduce to the earlier hyperbolic integral. Exchanging the two nodes, $s\leftrightarrow t$, reverses the displayed phase and leaves those factors and the measure unchanged. This explicitly recovers the original convention and hence $Z(1,k)=1/(4k)$.

There is a useful check for nonzero $M$. A real, convergent determinant difference is

$$\log\frac{D_b(M)}{D_b(0)} =\int_0^\infty\frac{dt}{t}\, \frac{\sinh(Q_bt/2)}{\sinh(bt)\sinh(b^{-1}t)} \bigl[\cos(2Mt)-1\bigr]. \tag{AB14.10}$$

It follows by pairing (AB14.7), using the integral for a logarithm, summing the two mode series, and subtracting the same expression at zero mass. Near $t=0$ the integrand tends to $-Q_bM^2$; at infinity it decays exponentially. This subtraction is ultraviolet finite, while the absolute value $D_b(0)$ still uses the fixed normalization. Since the kernel is positive, $0<D_b(M)\leq D_b(0)$ for real $M$.

At rank one set $x=s-t$, $y=(s+t)/2$ as before. The common-center phase has the opposite sign but gives the same normalized approximate delta. The continuous bounded test function is now $D_b(x+M)D_b(x-M)$. Applying the already justified Gaussian limit therefore gives

$$\boxed{Z_b(1,k;M)=\frac{D_b(M)^2}{k}.} \tag{AB14.11}$$

This is a finite-rank result in the declared source prescription. In particular,

$$Z_1(1,k;M)=\frac1{4k\cosh^2\pi M}, \qquad \left.\frac{\partial^2F}{\partial M^2}\right|_{b=1,M=0} =2\pi^2,\qquad F=-\log Z. \tag{AB14.12}$$

The derivative follows from $F=\log(4k)+2\log\cosh\pi M$. It measures this normalized flavor-source direction; it is not automatically a stress-tensor coefficient or a derivative with respect to dimensional $\mu$. The latter has an extra $L^2$.

For general $b$, (AB14.10) gives another explicitly positive response, $F_{MM}(0)=8\int_0^\infty dt\,t\,\sinh(Q_bt/2)/[\sinh(bt)\sinh(b^{-1}t)]$. At $b=1$, expansion of $1/\sinh t$ into positive exponentials recovers $2\pi^2$. A source derivative follows the full multiplet, including $D_F(\chi)$, as in the contact-term lesson.

Finally, $b\leftrightarrow b^{-1}$ exchanges the two geometric circle scales and the two mode labels; the displayed integral is invariant. This is not an identification with every metric called a “squashed sphere.” Nor is a special complex mass a real mass: KNP’s later tuning (3.22), $m_b=i(b^2-3)/(4b)$ in its dual-theory variables, is generally imaginary and belongs to an analytic continuation with its own pole/contour conditions. None of that restricted large-rank analysis was needed here.

Exercises

1. Change the flavor generator

Replace the chosen charges by $(+1,-1,0,0)$ on $(A_1,A_2,B_1,B_2)$, using the same real source $\mu$. Does the fixed superpotential permit this deformation? Write the four double-sine factors for one color pair with $x=s_i-t_j$. Must they still equal $D_b(x+M)D_b(x-M)$? Contrast the assignment $(+1,+1,+1,+1)$.

Hint 1

Each superpotential monomial contains one of each chiral. Test its total charge before changing a determinant.

Hint 2

A determinant depends on the actual gauge weight plus the assigned flavor mass. The $B$ fields now have no flavor shift.

Solution

The new charges sum to zero, so the fixed quartic coefficient remains neutral. Freeze the same complete multiplet (AB14.3). The four factors are

$$s_b(iQ_b/4-x-M)\, s_b(iQ_b/4-x+M)\, s_b(iQ_b/4+x)^2 .$$

They reduce to the original massless product at $M=0$, but their shifted partners no longer form the two $D_b$ factors of (AB14.8). They define a different flavor deformation. For example, at $x=0$ the product is $D_b(M)D_b(0)$, whereas the original charge choice gives $D_b(M)^2$; their nonzero mass responses already differ.

The all-positive assignment gives charge four to each superpotential word and is broken by a fixed nonzero quartic coefficient. Assigning a compensating charge to that coefficient as a formal spurion would compare different couplings; it does not make this an unbroken flavor symmetry of the fixed theory. A determinant shift alone cannot repair that problem.

2. An elementary answer away from the round sphere

At $b=\sqrt3$, use the shift relation

$$\frac{s_b(z+i/(2b))}{s_b(z-i/(2b))} =\frac1{2\cosh(\pi z/b)}$$

to simplify $D_b(M)$. Derive $Z_{\sqrt3}(1,k;M)$ and its mass Hessian at zero. Does its massless value still equal the round-sphere answer?

Hint 1

Here $Q_b/4=1/b$. Split the ratio from $M-i/b$ to $M+i/b$ into two half-step ratios through $M$.

Hint 2

Use $4\cosh(u+i\pi/6)\cosh(u-i\pi/6)=1+2\cosh(2u)$.

Solution

The two shift relations give

$$D_{\sqrt3}(M) =\frac1{4\cosh(\pi M/\sqrt3+i\pi/6) \cosh(\pi M/\sqrt3-i\pi/6)} =\frac1{1+2\cosh(2\pi M/\sqrt3)}.$$

This is the simplest non-round case of KNP (A.8). Consequently

$$Z_{\sqrt3}(1,k;M) =\frac1{k[1+2\cosh(2\pi M/\sqrt3)]^2}, \qquad F_{MM}(0)=\frac{16\pi^2}{9}.$$

At zero mass $Z=1/(9k)$ rather than $1/(4k)$. Supersymmetry has made the answer computable; it has not made it independent of the specified squashing source. The result is real and positive for real $M$ and has the expected evenness under $M\mapsto-M$.

3. Change a contact prescription

On the same compact flavor cover, add an integer-level background term

$$\Delta S_{\rm ct}=\frac{\kappa}{4\pi} \int_{S^3}\left(i a_F\wedge da_F-2\mu D_F\,d\mathrm{vol}\right), \qquad\kappa\in\mathbb Z.$$

Compute its effect on $Z_b$, $F_{MM}$ and $F_{\mu\mu}$ at fixed $L,b$. Does it change the real response in (AB14.12) or the masses in (AB14.4)?

Hint 1

Use the positive-$H$ source $D_F=-i\mu/f$ and the same volume integral that fixed the dynamical CS phase.

Hint 2

A background-only action adds directly to $F=-\log Z$. Distinguish a derivative in $M=L\mu$ from one in $\mu$.

Solution

Since $a_F=0$,

$$\Delta S_{\rm ct} =\frac{i\kappa\mu^2}{2\pi}\int\frac{d\mathrm{vol}}f =i\pi\kappa M^2 .$$

Therefore $Z_b$ is multiplied by $e^{-i\pi\kappa M^2}$, $\Delta F_{MM}=2i\pi\kappa$, and $\Delta F_{\mu\mu}=2i\pi\kappa L^2$. These are local contact changes in this compact-source normalization. They are imaginary for the real mass path, so the real Hessian is unchanged. The term contains no propagating matter field and changes neither quadratic operator. Its sign differs from the negative-$H$ round example because the auxiliary source has changed, not because the integer level was redefined.

A deformed integral now carries a definite physical specification: the gauge group and levels, complete geometry, flavor charges, auxiliary sources, dimensionless parameters and contact prescription. These are the data needed before comparing exact answers or interpreting their derivatives.

Continue with fourth mass responses and fixed-source squashing derivatives to see how nonlinear source terms enter integrated CFT data.